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Question

If x2+y2+z2=1, then yz+zx+xy lies in the interval

A
[12,2]
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B
[1,2]
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C
[12,1]
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D
[1,12]
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Solution

The correct option is C [12,1]
x2+y2+z2=1 [1]
(xy)2+(yz)2+(zx)2=2(x2+y2+z2xyyzzx)
(xy)2+(yz)2+(zx)2=2[1(xy+yz+zx)]
xy+yz+zx=112[(xy)2+(yz)2+(zx)2]
xy+yz+zx1 [2]
As [(xy)2+(yz)2+(zx)20]

Again,
(x+y+z)2=x2+y2+z2+2(xy+yz+zx)
(x+y+z)2=1+2(xy+yz+zx)
xy+yz+zx=12[(x+y+z)21]
xy+yz+zx12 [3]
[(x+y+z)20]

From [2] and [3],
xy+yz+zx[12,1]

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